package algorithm.problems.dynamic_programming;

import java.util.Arrays;

/**
 * Created by gouthamvidyapradhan on 27/01/2018.
 * In a N x N grid representing a field of cherries, each cell is one of three possible integers.

 0 means the cell is empty, so you can pass through;
 1 means the cell contains a cherry, that you can pick up and pass through;
 -1 means the cell contains a thorn that blocks your way.
 Your task is to collect maximum number of cherries possible by following the rules below:

 Starting at the position (0, 0) and reaching (N-1, N-1) by moving right or down through valid path cells (cells
 with value 0 or 1);
 After reaching (N-1, N-1), returning to (0, 0) by moving left or up through valid path cells;
 When passing through a path cell containing a cherry, you pick it up and the cell becomes an empty cell (0);
 If there is no valid path between (0, 0) and (N-1, N-1), then no cherries can be collected.
 Example 1:
 Input: grid =
 [[0, 1, -1],
 [1, 0, -1],
 [1, 1,  1]]
 Output: 5
 Explanation:
 The player started at (0, 0) and went down, down, right right to reach (2, 2).
 4 cherries were picked up during this single trip, and the matrix becomes [[0,1,-1],[0,0,-1],[0,0,0]].
 Then, the player went left, up, up, left to return home, picking up one more cherry.
 The total number of cherries picked up is 5, and this is the maximum possible.
 Note:

 grid is an N by N 2D array, with 1 <= N <= 50.
 Each grid[i][j] is an integer in the set {-1, 0, 1}.
 It is guaranteed that grid[0][0] and grid[N-1][N-1] are not -1.

 Solution O(N ^ 3) time-complexity. Traversing from (0, 0) -> (n - 1, n - 1) or the other way around both are the same.
 The key point to note here is the traversal should be done by two person simultaneously starting from (0, 0).
 Notice after t steps, each position (r, c) we could be, is on the line r + c = t (along the diagonal). So if we have
 two people at positions (r1, c1) and (r2, c2), then r2 = r1 + c1 - c2.

 */
public class CherryPickup {

    /**
     * Main method
     * @param args
     * @throws Exception
     */
    public static void main(String[] args) throws Exception{
        int[][] A = {{0 ,1,-1},{1, 0,-1},{1,1,1}};
        System.out.println(new CherryPickup().cherryPickup(A));
    }

    public int cherryPickup(int[][] grid) {
        int[][][] DP = new int[grid.length][grid.length][grid.length];
        for(int i = 0; i < grid.length; i ++){
            for(int j = 0; j < grid.length; j ++){
                Arrays.fill(DP[i][j], -1);
            }
        }
        int result = dp(grid.length, 0, 0, 0, DP, grid);
        if(result < 0) return 0;
        else return result;
    }

    private int dp(int N, int r1, int c1, int c2, int[][][] DP, int[][] grid){
        int r2 = r1 + (c1 - c2);
        if(r1 >= N || c1 >= N || c2 >= N || r2 >= N || grid[r1][c1] == -1 || grid[r2][c2] == -1) return Integer.MIN_VALUE;
        else if(DP[r1][c1][c2] != -1) return DP[r1][c1][c2];
        else if(r1 == N - 1 && c1 == N - 1) return grid[N - 1][N - 1];
        else{
            int max = (c1 == c2) ? grid[r1][c1] : (grid[r1][c1] + grid[r2][c2]);
            //verify all the 4 possibilities. (P1 down + P2 right), (P1 right, P2 right), (P1 right + P2 down),
            // (P1 down, P2 down)
            max += Math.max(Math.max(Math.max(dp(N, r1 + 1, c1, c2, DP, grid), dp(N, r1 + 1, c1, c2 + 1, DP, grid)), dp(N,
                    r1, c1 + 1, c2, DP, grid)), dp(N, r1, c1 + 1, c2 + 1, DP, grid));
            DP[r1][c1][c2] = max;
            return max;
        }
    }

}
